Unit 3 The Relational Model And From ER Diagram To Relational … · 2019-05-07 · The given ER diagram is clear, other than • Discovered, which is the continent in which a particular

© 2014 Zvi M. Kedem 1

Unit 3 The Relational Model

And From ER Diagram To Relational Database


Sets, Relations, and Tables

◆  In this unit, we learn the semantics of specifying a relational database, later we will learn the syntax of SQL for doing this

◆  The basic “datatype”, or “variable” of a relational database is a relation

◆  In this unit, such a variable will be a set ◆  Later, we will extend this, and such a variable will be a

multiset ◆  In SQL, such a variable is called a table ◆  We may use the term table for a relation in this unit too ◆  Good review of basic concepts and terminology is

available at http://en.wikipedia.org/wiki/Relational_databases


Sets

◆  We will not use axiomatic set theory ◆  A set is a “bag” of elements, some/all of which could be

sets themselves and a binary relationship “is element of” denoted by ∈, such as 2 ∈ {2, 5, 3, 7}, {2,8} ∈ {2, {2, 8}, 5, 3, 7},

◆  You cannot specify •  How many times an element appears in a set (if you could, this

would be a multiset) •  In which position an element appears (if you could, this would be a

sequence) ◆  Therefore, as sets: {2, 5, 3, 7} = {2, 7, 5, 3, 5, 3, 3} ◆  Note: in many places you will read: “an element can

appear in a set only once” This is not quite right. And it is important not to assume this, as we will see in a later unit


Sets

◆  Two sets A and B are equal iff (if and only if) they have the same elements

◆  In other words, for every x: x is an element of A iff (if and only if) x is an element of B

◆  In still different way: A and B are equal iff for every possible x, the questions “is x in A” and “is x in B” give the same answer •  Note there is no discussion or even a way of saying how many

times x appears in a set, really either none or at least once

◆  “More mathematically,” A = B means ∀ x { x ∈ A ↔ x ∈ B }

◆  Therefore, as sets: {2, 5, 3, 7} = {2, 7, 5, 3, 5, 3, 3} ◆  This reiterates what we have said previously


Relation

◆  Consider a table, with a fixed number of columns where elements of each column are drawn from some specific domain

◆  The columns are labeled and the labels are distinct ◆  We will consider such a table to be a set of rows (another

word for “row”: tuple) ◆  An example of a table S of two columns A and B

◆  A relation is such a table ◆  We will also write S(A,B) for table S with columns A and B

S A B a 2 a 2 b 3 c 4 d 3


Relational Schema

◆  What we saw was an instance (current value for a relation with the defined columns and domains)

◆  To specify this relation in general (not the specific instance) we need to talk about a relational schema

◆  A relational schema defines a set of relations ◆  In databases everything is finite, so a relational schema

defines a finite set of finite relations


Relational Schema

◆  Here is an informal, but complete, description what is a relational schema of one relation

◆  We want to define a structure for some table 1.  We give it a name (we had S) 2.  We chose the number of columns (we had 2) and give

them distinct names (we had A and B) 3.  We decide on the domains of elements in the columns

(we had letters for A and integers for B) 4.  We decide on constraints, if any, on the permitted values

(for example, we can assume as it was true for our example that any two rows that are equal on A must be equal on B)


Relational Schema

◆  Let’s verify •  A: all lower case letters in English •  B: all positive integers less than 100 •  S(A,B) satisfies the condition that any two tuples that are equal on

A must also be equal on B

◆  Our example was an instance of this relational schema

S A B a 2 a 2 b 3 c 4 d 3


Relations

◆  Since relations are sets of tuples, the following two relations are equal (are really one relation written in two different ways) (This is a different example, not an instance of the previous relational schema)

S A B a 2 a 56 b 2

S A B a 56 a 2 b 2 a 56 a 2 a 56


Relations

◆  Since the positions in the tuple (1st, 2nd, etc.) are labeled with the column headings, the following two relations are equal (are really one relation written in two different ways)

S A B a 2 a 56 b 2

S B A 56 a

2 a 2 b

56 a 2 a

56 a


Relations

◆  To specify relations, it is enough to do what we have done above

◆  As long as we understand what are the domains for the columns, the following are formally fully specified relations (recall Unit 1) •  Relational (schema) P(Name, SSN, DOB, Grade) with some (not

specified, but we should have done it) domains for attributes •  Relational (schema) Q(Grade, Salary) with some (not specified,

but we should have done it) domains for attributes

P Name SSN DOB Grade

A 121 2367 2

B 101 3498 4

C 106 2987 2

Q Grade Salary

1 90

2 80

3 70

4 70


Relations

◆  But we will do more. We will specify, as suitable for the schema: •  Primary keys •  Keys (beyond primary) •  Foreign keys and what they reference (we will see soon what this

means) •  Additional constraints

◆  Some of the constraints involve more than one relation ◆  The above are most important structurally ◆  Later, when we talk about SQL DDL, we will specify

additional properties


Keys (and Superkeys)

◆  Consider relation (schema) Person(FN, LN, Grade, YOB) ◆  Instance:

◆  We are told that any two tuples that are equal on both FN

and LN are (completely) equal •  We have some tuples appearing multiple times: this is just for

clarifying that this permitted in the definition, we do not discuss here why we would have the same tuple more than one time (we will talk about this later)

◆  This is a property of every possible instance of Person in our application—we are told that

◆  Then (FN, LN) is a superkey of Person, and in fact a key, because neither FN nor LN by themselves are sufficient (we are told that too)

Person FN LN Grade YOB John Smith 8 1976 Lakshmi Smith 9 1981 John Smith 8 1976 John Yao 9 1992



◆  Consider relation (schema) Q(Grade, Salary) ◆  Example:

◆  We are told that for any instance of Pay, any two tuples that are equal on Grade are (completely) equal •  Of course, if each Grade appears in only one tuple, this is

automatically true ◆  Then, similarly to before, Grade is a key ◆  What about Salary, is this a key also? ◆  No, because we are not told (that is, we are not

guaranteed) that any two tuples that are equal on Salary are equal on Grade in every instance of Pay

Pay Grade Salary

8 128

9 139

7 147



◆  A set of columns in a relation is a superkey if and only any two tuples that are equal on the elements of these columns are (completely equal)

◆  A relation always has at least one superkey ◆  The set of all the attributes is a superkey ◆  Because any two tuples that are equal on all attributes are

completely equal ◆  A minimal superkey, is a key ◆  A relation always has at least one key (start with any

superkey and remove unnecessary columns) ◆  There may be more than one key ◆  Exactly one key is chosen as primary key ◆  Other keys are just keys ◆  Sometimes they are called candidate keys (as they are

candidates for the primary key, though not chosen)



◆  To summarize

◆  Superkey: a set of columns whose values determine the values of all the columns in a row for every instance of the relation •  All the columns are a superkey (trivially)

◆  Key: a set of columns whose values determine the values of all the columns in a row for every instance of the relation, but any proper subset of the columns does not do that

◆  Primary Key: a chosen key



◆  We will underline the attributes of the chosen primary key ◆  Returning to Unit 2 and example of City:

City(Longitude,Latitude,Country,State,Name,Size) ◆  We can have

•  City(Longitude,Latitude,Country,State,Name,Size) •  This implies that Longitude,Latitude form a primary key •  We also have a candidate key: Country,State,Name

◆  We can have •  City(Longitude,Latitude,Country,State,Name,Size) •  This implies that Country,State,Name form a primary key •  We also have a candidate key: Longitude,Latitude


Relational Databases

◆  A relational database is “basically” a set of relations ◆  It is an instance of a relational schema

◆  This is what is formally correct, but it is misleading in practice

◆  As we will see later, a relational database is •  A set of relations •  A set of binary, many-to-one mappings between them (partial

functions)

◆  We will know later what this means exactly, but I did not want to leave you with not quite a useful definition


From ER Diagrams To Relational Database

◆  We are now ready to convert ER diagrams into relational databases

◆  Generally, but not always •  An entity set is converted into a table •  A relationship is converted into a table

◆  We will first go through a simple example ◆  Then, we will go through our large example, studied

previously ◆  Then, we look at some additional points of interest ◆  Finally, we summarize the process, so we are sure we

understand it


Small ER Diagram

Employee

Likes

Country

Born

Name Population

Animal

Species Discovered

NameID# Child


More About The Example

◆  The given ER diagram is clear, other than •  Discovered, which is the continent in which a particular species

was first discovered

◆  Each child is a “dependent” of only one employee in our database •  If both parents are employees, the child is “assigned” to one of

them

◆  We are given additional information about the application •  Values of attributes in a primary key must not be missing (this

is a general rule, not only for this example) •  Other than attributes in a primary key, other attributes, unless

stated otherwise, may be missing •  The value of Name is known (not missing) for every Employee

◆  To build up our intuition, let’s look at some specific instance of our application


Small Example

◆  5 Employees: •  1 is Alice has Erica

and Frank, born in US, likes Horse and Cat

•  2 is Bob has Bob and Frank, born in US, likes Cat

•  4 is Carol •  5 is David, born in IN •  6 is Bob, born in CN,

likes Yak

◆  4 Countries •  US •  IN has 1150 •  CN has 1330 •  RU

◆  4 Animals •  Horse in Asia •  Wolf in Asia •  Cat in Africa •  Yak in Asia •  Zebra in Africa


Country

◆  There are four countries, listing for them: Cname, Population (the latter only when known): •  US •  IN, 1150 •  CN, 1330 •  RU

◆  We create a table for Country “in the most obvious way,” by creating a column for each attribute (underlining the attributes of the primary key) and this works:

◆  Note that some “slots” are NULL, indicated by emptiness

Country Cname Population

US

IN 1150

CN 1330

RU


Animal

◆  There are five animals, listing for them: Species, Discovered (note, that even though not required, Discovered happens to be known for every Species): •  Horse, Asia •  Wolf, Asia •  Cat, Africa •  Yak, Asia •  Zebra, Africa

◆  We create a table for Animal as before, and this works:

Animal Species Discovered

Horse Asia

Wolf Asia

Cat Africa

Yak Asia

Zebra Africa


Employee

◆  There are five employees, listing for them: ID#, Name, (name of) Child (note there may be any number of Child values for an Employee, zero or more): •  1, Alice, Erica, Frank •  2, Bob, Bob, Frank •  4, Carol •  5, David •  6, Bob, Frank

◆  We create a table for Employee in the most obvious way, and this does not work:

Employee ID# Name Child Child

1 Alice Erica Frank

2 Bob Bob Frank

4 Carol

5 David

6 Bob Frank


Employee

◆  Child is a multivalued attribute so, the number of columns labeled “Child” is, in principle, unbounded

◆  A table must have a fixed number of columns •  It must be an instance in/of a relational schema

◆  If we are ready to store up to 25 children for an employee and create a table with 25 columns for children, perhaps tomorrow we get an employee with 26 children, who will not “fit”

◆  We replace our attempted single table for Employee by two tables •  One for all the attributes of Employee other than the multivalued

one (Child) •  One for pairs of the form (primary key of Employee, Child)

◆  Note that both tables have a fixed number of columns, no matter how many children an employee has


Employee And Child

◆  Replace (incorrect)

By (correct)


1 Alice Erica Frank

2 Bob Bob Frank

4 Carol

5 David

6 Bob Frank

Employee ID# Name

1 Alice

2 Bob

4 Carol

5 David

6 Bob

Child ID# Child

1 Erica

1 Frank

2 Bob

2 Frank

6 Frank


Employee And Child With Better Column Names

◆  Replace (incorrect)

By (correct)


1 Alice Erica Frank

2 Bob Bob Frank

4 Carol

5 David

6 Bob Frank

Employee ID# Name

1 Alice

2 Bob

4 Carol

5 David

6 Bob

Child Parent Child

1 Erica

1 Frank

2 Bob

2 Frank

6 Frank


Employee And Child

◆  The primary key of the table Employee is ID# ◆  The primary key of the table Child is the pair: ID#,Child ◆  One attribute is not sufficient to get a primary key for Child

◆  It is clear from the example how to handle any number of multivalued attributes an entity has •  Create a “main” table with all the attributes other than

multivalued ones Its primary key is the original primary key of the entity set

•  Create a table for each multivalued attribute consisting a primary key for the main table and that multivalued attribute

Its primary key is the primary key of the entity combined with the multivalued attribute


Foreign Key

◆  Let us return to our example ◆  Note that any value of ID# that appears in Child must also

appear in Employee •  Because a child must be a dependent of an existing employee

◆  This is an instance of a foreign key ◆  ID# in Child is a foreign key referencing Employee

•  This means that ID# appearing in Child must appear in some row “under” columns (here only one) of primary key in Employee

•  Note that ID# is not a key of Child (but is part of a key), so a foreign key in a table does not have to be a key of that table

Employee ID# Name

1 Alice

2 Bob

4 Carol

5 David

6 Bob

Child ID# Child

1 Erica

1 Frank

2 Bob

2 Frank

6 Frank


Foreign Key ≡ A Binary Many-To-One Relationship Between Tables (Partial Function)

◆  Note: •  Every row of Child has a single value of a primary key of

Employee, so every row of Child “maps” to a single row of Employee

•  Every row of Employee has zero or more rows of Child mapped into it

In other words, no constraint


Foreign Key ≡ A Binary Many-To-One Relationship Between Tables

◆  Another option ◆  Note names do not have to be the same for the mapping

to take place ◆  But you need to specify which column is foreign key

referring to what •  Here: Parent in Child is foreign key referencing Employee

Employee ID# Name

1 Alice

2 Bob

4 Carol

5 David

6 Bob

Child Parent Child

1 Erica

1 Frank

2 Bob

2 Frank

6 Frank


Born

◆  Born needs to specify which employees were born in which countries (for whom this information is known)

◆  We can list what is the current state •  Employee identified by 1 was born in country identified by US •  Employee identified by 2 was born in country identified by IN •  Employee identified by 5 was born in country identified by IN •  Employee identified by 6 was born in country identified by CN


Born

◆  Born needs to specify who was born where ◆  We have tables for

•  Employee •  Country

◆  We know that each employee was born in at most one country (actually was born in exactly one country but we may not know what it is)

◆  We have a binary many-to-one relationship between Employee and Country

Employee ID# Name

1 Alice

2 Bob

4 Carol

5 David

6 Bob


US

IN 1150

CN 1330

RU


Implementation For Born

◆  Augment Employee so instead of

we have

Employee ID# Name

1 Alice

2 Bob

4 Carol

5 David

6 Bob

Employee ID# Name Cname

1 Alice US

2 Bob IN

4 Carol

5 David IN

6 Bob CN


US

IN 1150

CN 1330

RU


US

IN 1150

CN 1330

RU


Implementation For Born

◆  Augment Employee so instead of

we have two tables and a binary many-to-one mapping

Employee ID# Name

1 Alice

2 Bob

4 Carol

5 David

6 Bob


1 Alice US

2 Bob IN

4 Carol

5 David IN

6 Bob CN


US

IN 1150

CN 1330

RU


US

IN 1150

CN 1330

RU


Foreign Key Constraint Implementing Born

◆  We have again a foreign key constraint ◆  Any value of Cname in Employee must also appear in

Country as a primary key in some row ◆  Cname in Employee is a foreign key referencing Country

◆  Note that Cname in Employee is not even a part of its primary key


1 Alice US

2 Bob IN

4 Carol

5 David IN

6 Bob CN


US

IN 1150

CN 1330

RU


Foreign Key Constraint Implementing Born

◆  Perhaps better (and frequently done in practice) use a different name for foreign keys

◆  Any value of CBirth in Employee must also appear in Country as a primary key in some row

◆  CBirth in Employee is a foreign key referencing Country

◆  We will not talk about such, possibly convenient, renaming

Employee ID# Name CBirth

1 Alice US

2 Bob IN

4 Carol

5 David IN

6 Bob CN


US

IN 1150

CN 1330

RU


Likes

◆  Likes needs to specify which employees like which animals

◆  We can list what is the current state: •  Employee identified by 1 likes animal identified by Horse •  Employee identified by 1 likes animal identified by Cat •  Employee identified by 2 likes animal identified by Cat •  Employee identified by 6 likes animal identified by Yak


Likes

◆  We can describe Likes by drawing lines between the two tables

◆  We need to “store” this set of red lines ◆  Likes is a many-to-many relationship

•  It is not a many-to-one relationship and therefore it is not a partial function


Horse Asia

Wolf Asia

Cat Africa

Yak Asia

Zebra Africa

Employee ID# Name

1 Alice

2 Bob

4 Carol

5 David

6 Bob


Likes (impossible implementation)

◆  Cannot store with Employee (there is no limit on the number of animals an employee likes)


Horse Asia

Wolf Asia

Cat Africa

Yak Asia

Zebra Africa

Employee ID# Name

1 Alice

2 Bob

4 Carol

5 David

6 Bob

Employee ID# Name Species Species

1 Alice Horse Cat

2 Bob Cat

4 Carol

5 David

6 Bob Yak


Likes (impossible implementation)

◆  Cannot store with Animal (there is no limit on the number of employees who like an animal)


Horse Asia

Wolf Asia

Cat Africa

Yak Asia

Zebra Africa

Employee ID# Name

1 Alice

2 Bob

4 Carol

5 David

6 Bob

Animal Species Discovered ID# ID#

Horse Asia 1

Wolf Asia

Cat Africa 1 2

Yak Asia 6

Zebra Africa


Likes

◆  Each red line is an edge defined by its vertices ◆  We create a table storing the red lines; that is, its vertices ◆  We can do this using the primary keys of the entities

We do not need other attributes such as Name or Discovered

◆  The table for Likes contains tuples: •  1 likes Horse •  1 likes Cat •  2 likes Cat •  6 likes Yak

Likes ID# Species

1 Horse

1 Cat

2 Cat

6 Yak


Likes

◆  Note that there are foreign key constraints •  ID# appearing in Likes is a foreign key referencing Employee •  Species appearing in Likes is a foreign key referencing Animal

◆  And two many-to-one mappings are induced ◆  Note: a binary many-to-many relationship was

replaced by a new table and two many-to one relationships


Using Visio

◆  We will use Visio for designing/specifying relational databases

◆  You can look at a tutorial, to get familiar with the mechanics of Visio

◆  This is greatly oversimplified, but a good start •  http://www.youtube.com/watch?v=1BYt3wmkgXE but foreign keys

are not explained •  http://www.youtube.com/watch?v=55TpWp4TmMw&NR=1 •  http://www.youtube.com/watch?v=r0x8ZMyPoj4&NR=1 but this

third part –  Is misleading in the context of relational databases, due to the handling of

many-to-many relationships and –  They use of the second page, all the pages in a single Visio drawing refer to a

single ER diagram, so each ER diagram needs its own Visio drawing/file


Specifying A Relational Implementation

◆  We will use Visio to specify our relational implementation ◆  And in fact, we could even use software to generate

database specifications from the diagram to SQL DDL ◆  We will just focus for now on the first task


Specifying A Relational Implementation Using Visio 2010

◆  A drawing in Visio is not an Entity Relationship Diagram tool despite such terminology in Visio

◆  This is good, as it produces a relational schema, which is what we actually need, but this is a lower-level construct

◆  It focuses on tables and the implicit many-to-one binary relationships induced by foreign key constraints

◆  Table •  A rectangle with three vertical subrectangles: name, list of

attributes in the primary key, list of attributes not in the primary key •  Required attributes are in bold •  Attributes in the primary key and foreign keys are labeled as such

◆  Relationship •  A many-to-one binary (or perhaps one-to-one, which is a special

case) relationship induced by a foreign key constraint is explicitly drawn by means of a segment with an arrow head

We will have alternative notations later


Relational Implementation For The Example

Child ID# Child

1 Erica

1 Frank

2 Bob

2 Frank

6 Frank

Employee ID# Name CName

1 Alice US

2 Bob IN

4 Carol

5 David IN

6 Bob CN

Likes ID# Species

1 Horse

1 Cat

2 Cat

6 Yak


Horse Asia

Wolf Asia

Cat Africa

Yak Asia

Zebra Africa

Country CName Population

US

IN 1150

CN 1330

RU

Employee

PK ID#

NameFK1 CName

Child

PK,FK1 ID#PK Child

Country

PK CName

Population

Likes

PK,FK1 ID#PK,FK2 Species

Animal

PK Species

Discovered


Cardinality Constraints

◆  The statement that a relationship is many-to-one as opposed to be a “standard” many-to-many relationship is really a cardinality constraint

◆  We will look at a relationships Likes between Person and Country and four cases of cardinality constraints on how many Countries a Person may like •  No constraint •  At least one •  At most one •  Exactly one

◆  For the first two, Likes is many-to-many ◆  For the last two, Likes is many-to-one ◆  Intuitively, Likes is many to one if for every Person, when

you see which Countries this Person Likes, you get 0 or 1 ◆  If you always get 1, this is a total function, otherwise this is

a partial function


Specifying These Constraints (Revisited From Unit 2)

0 ..*

1 ..*

0 .. 1

1 .. 1

Person

Person

Person

Person

Likes

Likes

Likes

Likes

Country

Country

Country

Country

Every Person likes 0 or more Countries

Every Person likes 1 or more Countries

Every Person likes 0 or 1 Countries

Every Person likes 1 Country


Arrow Notation Cannot Distinguish Some Cases

0 ..*

1 ..*

0 .. 1

1 .. 1

Person

Person

Person

Person

Likes

Likes

Likes

Likes

Country

Country

Country

Country

Person

Person

Person

Person

Likes

Likes

Likes

Likes

Country

Country

Country

Country


Crow’s Feet: Improved Arrow Notation

◆  Note: different sides of the relationship are labeled in the two notations!

0 ..*

1 ..*

0 .. 1

1 .. 1

LikesPerson

Person

Person

Person

Likes

Likes

Likes

Likes

Country

Country

Country

Country

Person

Person

Person

Person

Country

Country

Country

Country

Likes

Likes

Likes


Crow’s Feet

◆  In general, cardinalities of both sides of the relationship may need to be specified

◆  We did only one, because it is sufficient to understand the notation

◆  We now return to the relational implementation of our example

◆  Visio can use the Crow’s Feet notation


Relational Implementation For The Example

Employee

PK ID#

NameFK1 CName

Child

PK,FK1 ID#PK Child

Country

PK CName

Population

Likes


Animal

PK Species

Discovered

Likes ID# Species

1 Horse

1 Cat

2 Cat

6 Yak


Horse Asia

Wolf Asia

Cat Africa

Yak Asia

Zebra Africa

Country CName Population

US

IN 1150

CN 1330

RU

Child ID# Child

1 Erica

1 Frank

2 Bob

2 Frank

6 Frank

Employee ID# Name CName

1 Alice US

2 Bob IN

4 Carol

5 David IN

6 Bob CN


End Of Lines In Crow’s Feet Notation

◆  0..1

◆  0..*

◆  1..1

◆  1..*


Intuition For The Notation: One-Level Tree

11 6633 88 99

AA

Many Side

One Side


Pattern Of Lines

◆  The line between Animal and Likes is solid because the primary key of the “many side”, Likes, includes the primary key of the “one side”, Animal, so it “cannot exist” without it

◆  The line between Employee and Likes is solid because the primary key of the “many side”, Likes, includes the primary key of the “one side”, Employee, so it “cannot exist” without it

◆  The line between Employee and Child is solid because the primary key of the “many side”, Child, includes the primary key of the “one side”, Employee, so it “cannot exist” without it

◆  The line between Country and Employee is dashed because the primary key of the “many side”, Employee, does not include the primary key of the “one side”, Country, so it “can exist” without it


Pattern Of Lines

◆  This is not a question of the ends of lines “forcing” the pattern of lines

◆  In the next slide, we see a slight modification of our example in which all lines have the same pair of endings

◆  We required that for each Employee the Country of Birth is known

◆  Nevertheless, as Cname is not part of the primary key of Country, the line is dashed

◆  For technical reasons, the tables have slightly different names, but this has nothing to do with our point


Example

◆  Assume: Every employee has exactly one Country (that is we know the country of birth)

Employee1

PK ID#

NameFK1 CName

Child1

PK,FK1 ID#PK Child

Country1

PK CName

Population

Likes1


Animal1

PK Species

Discovered


Alternative Implementation For Born

◆  We need an “in-between” table for Likes because it is many-to-many

◆  We do not need an “in-between” table for Born because it is many-to-one

◆  But we can implement Born using such an “in-between” table


Alternative Implementation For The Example

Child ID# Child

1 Erica

1 Frank

2 Bob

2 Frank

6 Frank

Employee ID# Name

1 Alice

2 Bob

4 Carol

5 David

6 Bob

Born ID# CName

1 US

2 IN

5 IN

6 CN


US

IN 1150

CN 1330

RU

Likes ID# Species

1 Horse

1 Cat

2 Cat

6 Yak


Horse Asia

Wolf Asia

Cat Africa

Yak Asia

Zebra Africa

Employee

PK ID#

Name

Child

PK,FK1 ID#PK Child

Country

PK CName

Population

Likes


Animal

PK Species

Discovered

Born

PK,FK1 ID#

FK2 CName


Alternative Implementation For The Example

Child ID# Child

1 Erica

1 Frank

2 Bob

2 Frank

6 Frank

Employee ID# Name

1 Alice

2 Bob

4 Carol

5 David

6 Bob

Born ID# CName

1 US

2 IN

5 IN

6 CN


US

IN 1150

CN 1330

RU

Likes ID# Species

1 Horse

1 Cat

2 Cat

6 Yak


Horse Asia

Wolf Asia

Cat Africa

Yak Asia

Zebra Africa

Employee

PK ID#

Name

Child

PK,FK1 ID#PK Child

Country

PK CName

Population

Likes


Animal

PK Species

Discovered

Born

PK,FK1 ID#

FK2 CName


Options For Relationships

General case ◆  We have a relationship R among entity sets E1, E2, …, En,

with properties P1, P2, …, Pm

◆  Each Ei is implemented as a table ◆  We can always implement R as a table with foreign key

constraints referencing E1, E2, …, En, and with R also storing properties P1, P2, …, Pm

Special case: R is binary many-to-one from E1 to E2

◆  We can, if we like, avoid introducing a table for R ◆  We implement R as a foreign key constraint in E1

referencing E2, and with E1 also storing properties P1, P2, …, Pm


Which Implementation To Use For Born?

◆  We cannot give a general rule ◆  The first implementation uses more tables ◆  The second implementation may introduce NULLs (empty

values), which we do not like

◆  For the purpose of the class we will always use the second implementation, to have better exercises

◆  So do this for all the homeworks and tests, when relevant


To Remember!

◆  Structurally, a relational database consists of 1.  A set of tables with identifiers (primary keys) 2.  A set of many-to-one binary relationships between them,

induced by foreign key constraints In other words; a set of functions (in general partial), each from a table into a table

◆  When designing a relational database, you should specify both (or you will produce a bad specification) •  Technically, tables are enough, but this a very bad practice as you

do not specify the relationships between tables


Many-To-One Mapping From Child To Employee

◆  Partial mapping from a set of rows into a set of rows

1 Alice

2 Bob

4 Carol

5 David

6 Bob

1 Erica

1 Frank

2 Bob

2 Frank

6 Frank


Very Bad Visio Diagram

◆  Tables are listed with attributes, specifying only which are in the primary key

◆  Foreign key constraints are not specified •  So the DB system does not know what to enforce

Employee

PK ID#

Name CName

Child

PK ID#PK Child

Country

PK CName

Population

Likes

PK ID#PK Species

Animal

PK Species

Discovered


Terrible Visio Diagram

◆  Even primary keys are not specified

Employee

ID# Name CName

Child

ID# Child

Country

CName Population

Likes

ID# Species

Animal

Species Discovered


From ER Diagram To Relational Database

◆  We now convert our big ER diagram into a relational database

◆  We specify •  Attributes that must not be NULL •  Primary keys •  Keys (beyond primary) •  Foreign keys and what they reference •  Cardinality constraints •  Some additional “stubs”

◆  We both give a narrative description, similar to actual SQL DDL (so we are learning about actual relational databases) and Visio diagrams

◆  We should specify domains also, but we would not learn anything from this here, so we do not do it

◆  We go bottom up, in the same order as the one we used in constructing the ER diagram


Our ER Diagram

Name

FN LN

SS#ID# DOB AgeChild

PersonAutomobile

Model Year Weight

Likes

Student Professor

ISA SalaryGPAVIN Color

CarType Has0..1 2..*

SectionSec#

Year

Semester

TookGrade

MaxSize

TaughtMonitors

0..1

TitleC#

Offered1..1 1..*

3..50

Course

PrereqFirst Second

Book

Title Author

Required

1..1

Description

Horse

Name

Date


Hierarchy For Our ER Diagram

Car Automobile Person Course Book

Has Likes

Required

ProfessorISA

StudentISA

SectionOffered

TaughtTook

Prereq

Monitors

Horse

Note: circular dependency,

need to be treated together

Note: circular dependency,

need to be treated together

Note: circular dependency, need to be treated together

Type


We Will Produce

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child

PK,FK1 ID#PK ChildName

Automobile

PK ModelPK Year

Weight

Likes

PK,FK2 ID#PK,FK1 ModelPK,FK1 Year

Student

PK,FK1 ID#

Professor

PK,FK1 ID#

Salary

Course

PK C#

Title Description

Prereq

PK,FK2 FirstPK,FK1 Second

Book

PK AuthorPK Title

Required

PK,FK1 ID#PK,FK2 C#PK,FK3 AuthorPK,FK3 Title

Section

PK,FK1 C#PK YearPK SemesterPK Sec#

MaxSize

Took

PK,FK1 ID#PK,FK2 C#PK,FK2 YearPK,FK2 SemesterPK,FK2 Sec#

Grade

Taught


FK3 Monitor

Car

PK VIN

FK2 ID# ColorFK1 ModelFK1 Year Date


Horse

◆  Define Table Horse ( Name NOT NULL, Primary Key (Name));

◆  This represents the simplest possible relational database •  One table with one attribute


Horse

Horse

PK Name


Person

◆  Person has some interesting attributes ◆  Multivalued attribute: we will create another table ◆  Derived attribute: we do not create a column for it, it will

be computed as needed ◆  Composite attribute: we “flatten” it


Person

◆  Define Table Person ( ID# NOT NULL, SS# NOT NULL, FN, LN NOT NULL, DOB NOT NULL, Primary Key (ID#), Candidate Key (SS#), Age (computed by procedure …) );

◆  In SQL DDL, the keyword UNIQUE is used instead of Candidate Key, but “Candidate Key” is better for reminding us what this could be

◆  Age would likely not be stored but defined in some view


Person

Horse

PK Name

Person

PK ID#

SS# FN LN DOB


Child

◆  Define Table Child ( ID# NOT NULL, ChildName NOT NULL, Primary Key (ID#,ChildName), Foreign Key (ID#) References Person );

◆  This lists all pairs (ID# of person, a child’s name) •  We have chosen a more descriptive attribute name than the one in

the ER diagram for children’s names

◆  Note •  A person may have several children, each with a different name •  Two different persons may have children with the same name

◆  Because of this, no single attribute can serve as primary key of Child


Person And Child

◆  Note that some attributes are not bold, such as FN here ◆  This means that FN could be NULL (in this context,

meaning empty)

◆  Note the induced many-to-one relationship ◆  We need to make sure we understand what the line ends

indicate •  A person may have 0 or more children (unbounded) •  A child has exactly 1 person to whom it is attached

◆  We need to pay attention to such matters, though we are generally not going to be listing them here But you should look at all lines and understand the ends and the patterns (solid or dashed)


Person And Child

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child



Automobile

◆  Define Table Automobile ( Model NOT NULL, Year NOT NULL, Weight NOT NULL, Primary Key (Model,Year) );


Automobile

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight


Likes

◆  Define Table Likes ( ID# NOT NULL, Model NOT NULL, Year NOT NULL, Primary Key (ID#,Model,Year), Foreign Key (ID#) References Person, Foreign Key (Model,Year) References Automobile );

◆  Note: the following is bad/incorrect, replacing one line by two lines •  Foreign Key (Model) References Automobile •  Foreign Key (Year) References Automobile

◆  There are induced binary many-to-one relationships between •  Likes and Person •  Likes and Automobile


Likes

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes



Car

◆  Define Table Car ( VIN NOT NULL, Color, Primary Key (VIN) );


Car

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Car

PK VIN

Color


Type

◆  There is no need for a table for Type as Type is a binary many-to-one relationship

◆  It is essentially “stored” in the “many” side, that is in Car


Car

◆  Define Table Car ( VIN NOT NULL, Color, Model NOT NULL, Year NOT NULL, Primary Key (VIN), Foreign Key (Model,Year) References Automobile );


Type

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Car

PK VIN

ColorFK1 ModelFK1 Year


Has

◆  As Has is a binary many-to-one relationship, the attributed of this relationship, Date, is stored in the “many” side, Car

◆  There is no need for a table for Has as Has is a binary many-to-one relationship

◆  It is essentially “stored” in the “many” side, that is in Car

◆  We can only specify that a Person has at least 1 Car with the notation we currently use

◆  The CHECK condition is specified using appropriate SQL constraint syntax This can actually be done in Visio also, and it is done in the examples in ExtrasForUnit03


Car

◆  Define Table Car ( VIN NOT NULL, Color, Model NOT NULL, Year NOT NULL, ID#, Primary Key (VIN), Foreign Key (Model,Year) References Automobile Foreign Key (ID#) References Person );


Has

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Car

PK VIN



ISA

◆  We do not define a table for ISA ◆  This/these relationship/s is/are “embedded” in Student and

Professor


Student

◆  Define Table Student ( ID# NOT NULL, Primary Key (ID#), Foreign Key (ID#) References Person, GPA (computed by procedure …) );

◆  Note, how ISA, the class/subclass (set/subset) relations, is modeled by Visio


Student And ISA

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Student

PK,FK1 ID#

Car

PK VIN



Professor

◆  Define Table Professor ( ID# NOT NULL, Salary NOT NULL, Primary Key (ID#), Foreign Key (ID#) References Person );


Professor And ISA

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Student

PK,FK1 ID#

Professor

PK,FK1 ID#

Salary

Car

PK VIN



Course

◆  Define Table Course ( C# NOT NULL, Title NOT NULL, Description, Primary Key (C#) );


Course

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Student

PK,FK1 ID#

Professor

PK,FK1 ID#

Salary

Course

PK C#

Title Description

Car

PK VIN


© 2014 Zvi M. Kedem 100

Prerequsite

◆  Define Table Prereq ( First NOT NULL, Second NOT NULL, Primary Key (First,Second), Foreign Key (First) References Course, Foreign Key (Second) References Course );

© 2014 Zvi M. Kedem 101

Prereq

◆  This is our first example of a table modeling a recursive relationship, between an entity set and itself

◆  We decide to name the table Prereq, as this is shorter than Prerequisite

◆  Note that it is perfectly clear and acceptable to refer here to C# by new names: First and Second •  Similarly, to using ChildName in the Child table

◆  We should add some constraint to indicate that this (directed graph) should be acyclic (but as annotations) •  Maybe other conditions, based on numbering conventions

specifying course levels

© 2014 Zvi M. Kedem 102

Prereq

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Student

PK,FK1 ID#

Professor

PK,FK1 ID#

Salary

Course

PK C#

Title Description

Prereq


Car

PK VIN


© 2014 Zvi M. Kedem 103

Book

◆  Define Table Book ( Author NOT NULL, Title NOT NULL, Primary Key (Author,Title) );

© 2014 Zvi M. Kedem 104

Book

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Student

PK,FK1 ID#

Professor

PK,FK1 ID#

Salary

Course

PK C#

Title Description

Prereq


Book

PK AuthorPK Title

Car

PK VIN


© 2014 Zvi M. Kedem 105

Required

◆  Define Table Required ( ID# NOT NULL, C# NOT NULL, Author NOT NULL, Title NOT NULL, Primary Key (ID#,C#,Author,Title), Foreign Key (ID#) References Professor, Foreign Key (C#) References Course, Foreign Key (Author,Title) References Book );

◆  Why is it bad to have

Foreign Key (ID#) References Person, instead of Foreign Key (ID#) References Professor? Because only a Professor can Require a Book

© 2014 Zvi M. Kedem 106

Required

◆  This is our first example of a table modeling a relationship that is not binary

◆  Relationship Required was ternary: it involved three entity sets

◆  There is nothing unusual about handling it ◆  We still have as foreign keys the primary keys of the

“participating” entities

© 2014 Zvi M. Kedem 107

Required

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Student

PK,FK1 ID#

Professor

PK,FK1 ID#

Salary

Course

PK C#

Title Description

Prereq


Book

PK AuthorPK Title

Required


Car

PK VIN


© 2014 Zvi M. Kedem 108

Section

◆  Define Table Section ( C# NOT NULL, Year NOT NULL, Semester NOT NULL, Sec# NOT NULL, MaxSize, Primary Key (C#,Year,Semester,Sec#), Foreign Key (C#) References Course );

◆  Note on the end of the edge between Course and Section, the Section end, on the Visio drawing how the requirement of having at least one Section is modeled

© 2014 Zvi M. Kedem 109

Section

◆  Section is our first example of a weak entity

© 2014 Zvi M. Kedem 110

Offered

◆  We do not define a table for Offered

◆  Relationship Offered is implicit in the foreign key constraint

© 2014 Zvi M. Kedem 111

Section + Offered

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Student

PK,FK1 ID#

Professor

PK,FK1 ID#

Salary

Course

PK C#

Title Description

Prereq


Book

PK AuthorPK Title

Required


Section


MaxSize

Car

PK VIN


© 2014 Zvi M. Kedem 112

Took

◆  Define Table Took ( ID# NOT NULL, C# NOT NULL, Year NOT NULL, Semester NOT NULL, Sec# NOT NULL, Grade, Primary Key (ID#,C#,Year,Semester,Sec#), Foreign Key (ID#) References Student, Foreign Key (C#,Year,Semester, Sec#) References Section );

◆  Note on the end of the edge between Section and Took, the Took end, on the Visio drawing how the requirement of having between 3 and 50 students in a section is not fully modeled

◆  We can only show 1 or more using current notation

© 2014 Zvi M. Kedem 113

Took

◆  Because Took is a many-to-many relationship we store its attribute, Grade, in its table

◆  We cannot store Grade in any of the two •  Section •  Student

© 2014 Zvi M. Kedem 114

Took

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Student

PK,FK1 ID#

Professor

PK,FK1 ID#

Salary

Course

PK C#

Title Description

Prereq


Book

PK AuthorPK Title

Required


Section


MaxSize

Took


Grade

Taught


Car

PK VIN


© 2014 Zvi M. Kedem 115

Taught

◆  Define Table Taught ( ID# NOT NULL, C# NOT NULL, Year NOT NULL, Semester NOT NULL, Sec# NOT NULL, Primary Key (ID#,C#,Year,Semester,Sec#), Foreign Key (ID#), References Professor, Foreign Key (C#,Year,Semester,Sec#) References Section );

© 2014 Zvi M. Kedem 116

Taught

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Student

PK,FK1 ID#

Professor

PK,FK1 ID#

Salary

Course

PK C#

Title Description

Prereq


Book

PK AuthorPK Title

Required


Section


MaxSize

Took


Grade

Taught


Car

PK VIN


© 2014 Zvi M. Kedem 117

Monitors

◆  This is our first example in which a table, Taught, that “came from” a relationship is treated as if it came from an entity and participates in a relationship with other tables

◆  Nothing special needs to be done to “convert” a table that models a relationship, to be also treated as a table modeling an entity

◆  In this case, Monitors is a binary many-to-one relationship, so we do not need to create a table for it, and it can be stored in the “many” side, Taught

© 2014 Zvi M. Kedem 118

Taught

◆  Define Table Taught ( ID# NOT NULL, C# NOT NULL, Year NOT NULL, Semester NOT NULL, Sec# NOT NULL, Monitor Primary Key (ID#,C#,Year,Semester,Sec#), Foreign Key (ID#), References Professor, Foreign Key (C#,Year,Semester,Sec#) References Section Foreign Key (Monitor) References Professor );

© 2014 Zvi M. Kedem 119

Monitors

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Student

PK,FK1 ID#

Professor

PK,FK1 ID#

Salary

Course

PK C#

Title Description

Prereq


Book

PK AuthorPK Title

Required


Section


MaxSize

Took


Grade

Taught


FK3 Monitor

Car

PK VIN


© 2014 Zvi M. Kedem 120

We Are Done

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Student

PK,FK1 ID#

Professor

PK,FK1 ID#

Salary

Course

PK C#

Title Description

Prereq


Book

PK AuthorPK Title

Required


Section


MaxSize

Took


Grade

Taught


FK3 Monitor

Car

PK VIN


© 2014 Zvi M. Kedem 121

Arrow Notation

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Student

PK,FK1 ID#

Professor

PK,FK1 ID#

Salary

Course

PK C#

Title Description

Prereq


Book

PK AuthorPK Title

Required


Section


MaxSize

Took


Grade

Taught


FK3 Monitor

Car

PK VIN


© 2014 Zvi M. Kedem 122

Arrows And Cardinality Notation

Horse

PK Name

Person

PK ID#

SS# FN LN DOB

Child


Automobile

PK ModelPK Year

Weight

Likes


Student

PK,FK1 ID#

Professor

PK,FK1 ID#

Salary

Course

PK C#

Title Description

Prereq


Book

PK AuthorPK Title

Required


Section


MaxSize

Took


Grade

Taught


FK3 Monitor

Car

PK VIN


*

1..*

*

*

**

*

*

2*

*

*

3..50

*

*

*

© 2014 Zvi M. Kedem 123

Additional Points

◆  We did not write out on the diagram various constraints that must be known, such as •  At least preliminary domains, e.g., number, string, etc. •  What is the maximum permitted section size

◆  This must be done for proper documentation of the application’s requirements

◆  We will discuss some additional, important, points •  Elaboration on recursive relationships •  Referential Integrity •  Temporal databases

© 2014 Zvi M. Kedem 124

Recursive Relationships: Example

◆  Assume now that a prerequisite course, “First” course, must be taken with at least some Grade to count as a prerequisite

◆  This to make an example a little “richer” ◆  Two cases:

•  A course may have any number of prerequisites Prereq is many-to-many

•  A course may have at most one prerequisite Prereq is many to one (Second is the many side, a single First could be a prerequisite for many Second courses)

TitleC#

Course

PrereqFirst Second

Description TitleC#

Course1

PrereqFirst Second

Description

Grade Grade

© 2014 Zvi M. Kedem 125

Recursive Relationships: Example

◆  Nothing special, we handle the second case of Prereq by storing it in the “many” side of the relationship

◆  So there are two additional attributes in Course1 •  The prerequisite course, if any •  The required grade, if any

Course

PK C#

Title Description

Prereq


Grade

Course1

PK C#

Title DescriptionFK1 Prereq Grade

© 2014 Zvi M. Kedem 126

Referential Integrity: Example

◆  Assume that we have some professors in table Professor, with rows: 5,1 and 7,2

◆  There is a row in Taught 5,G22.2433,2009,Spring,001,7 ◆  This means that 5 teaches a specific section and 7

monitors this assignment

Taught ID# C# Year Semester Sec# Monitor

5 G22.2433 2009 Spring 001 7

Professor ID# Salary

5 1

7 2

© 2014 Zvi M. Kedem 127


◆  A user accesses the database and attempts to delete row (or all rows like this, recall that duplicates are permitted) 5,1 from Professor

◆  What should happen, as there is a row in Taught referencing this row in Professor?

◆  A user accesses the database and attempts to delete row 7,2 from Professor?

◆  What should happen, as there is a row in Taught referencing this row in Professor?

Taught ID# C# Year Semester Sec# Monitors

5 G22.2433 2009 Spring 001 7

Professor ID# Salary

5 1

7 2

© 2014 Zvi M. Kedem 128


◆  Part of specification of foreign key in in Taught ◆  An action on Professor can be denied, or can trigger an

action on Taught ◆  For example

•  ON DELETE NO ACTION This means that the “needed” row in Professor cannot be deleted Of course, it is possible to delete the row from Taught and then from the Professor (if no other row in in any table in the database “needs” the row in Professor)

•  ON DELETE CASCADE This means that if the a row is deleted from Professor, all the rows in Taught referring to it are deleted too

•  ON DELETE SET NULL This means, that the value referring to no-longer-existing professor is replaced by NULL

In our example, this is not possible for ID# as it is a part of the primary key of Taught, but is possible for Monitor

© 2014 Zvi M. Kedem 129

Referential Integrity: Another Example

◆  Part of specification of foreign key in in Professor ◆  An action on Person can be denied, or can trigger an

action on Professor ◆  For example

•  ON UPDATE CASCADE This means that if the value of ID# in Person is changed, this value of ID# also propagates to Professor

◆  Could (and probably should) add to Taught and Required: •  ON UPDATE CASCADE

In appropriate attributes, so that the change of ID# in Professor also propagates to them

In Taught in both ID# and Monitor In Required in ID#

◆  Excellent mechanism for centralized maintenance

© 2014 Zvi M. Kedem 130

Temporal Databases

◆  Of course, we may want to maintain historical data ◆  So, in practice one may have some indication that the

professor no longer works, but still keep historical information about the past

◆  But we do not assume this for our example

© 2014 Zvi M. Kedem 131

Summary: Strong Entity

◆  Example: Person ◆  Create a table for the entity without multivalued and

derived attributes, flattening composite attributes The primary key of this table will consist of the attributes serving as primary key of the entity Example table: Person

◆  If there is a derived attribute, describe how it is computed, but do not store it

◆  If there is a multivalued attribute, create a table for it consisting of it and attributes of the primary key of the entity; do not put it in the table for the entity Example table: Child The primary key of this table will consist of all its attributes

© 2014 Zvi M. Kedem 132

Summary: Strong Entity

◆  There could be an attribute that is composite with some components being multivalued and some derived

◆  And similar complexities ◆  Example, without drawing the appropriate entity using the

ER model (this is getting too hairy) •  A person has many children (multivalued) •  Each child has both FirstName and MiddleName •  The child has DOB •  The child has Age

◆  Then the table for child will look like

Child ID# FirstName MiddleName DOB

5432 Krishna Satya 2006-11-05

© 2014 Zvi M. Kedem 133

Summary: ISA And A Subclass

◆  Example: ISA and Professor ◆  Do not produce anything for ISA ◆  The class “above” ISA (here Person) has already been

implemented as a table ◆  Create a table with all the attributes of the subclass (as for

strong entity above) augmented with the primary key of the table “above” ISA, and no other attributes from it The primary key is the same as the primary key of the table “above” ISA Example table: Professor

© 2014 Zvi M. Kedem 134

Summary: Weak Entity And Defining Relationship

◆  Example: Offered and Section ◆  Do not produce anything for the defining relationship, here

Offered ◆  Imagine that the weak entity is augmented by the primary

key of the “stronger” table through which it is defined (the table for it has been created already) Treat the augmented weak entity the same way as a strong entity The primary key is the primary key of the “stronger” table augmented by the attributes in the discriminant of the weak entity (a discriminant may consist of more than one attribute) Example table: Section and Offered

© 2014 Zvi M. Kedem 135

Summary: A Relationship That Is Not Binary Many-To-One

◆  Example Took The tables for the participating entities have already been created Create a table consisting of the primary keys of the participating tables and the attributes of the relationship itself Of course, treat attributes of the relationship that are derived, multivalued, or composite, appropriately, not storing them, producing additional tables, flattening them The primary key consists of all the attributes of the primary keys of the participating tables Example table: Took

© 2014 Zvi M. Kedem 136

Summary: A Relationship That Is Binary Many-To-One

◆  Example: Has Do not create a table for this relationship Put the attributes of the primary key of the “one” side and the attributes of the relationship itself into the table of the “many” side Of course, treat attributes of the relation that are derived, multivalued, or composite, appropriately, not storing them, producing additional tables, flattening them, as the case may be You may decide to treat such a relationship the way you treat a relationship that is not binary many to one (but not in our class) If the relationship is one-to-one, choose which side to treat as if it were “many” Example table: Has

© 2014 Zvi M. Kedem 137

Summary: Treating A Relationship As An Entity

◆  Example: Taught (before it was modified by removing Approved) We have a table for that was created when we treated it as a relationship We do not need to do anything else to this table Example table: Taught

© 2014 Zvi M. Kedem 138

Key Ideas

◆  Sets ◆  Relations and tables ◆  Relational schema ◆  Primary keys ◆  Implementing an ER diagram as a relational schema

(relational database) ◆  General implementation of strong entities ◆  Handling attributes of different types ◆  General implementation of relationships ◆  Possible special implementation of binary many-to-one

relationships ◆  Implementation of ISA ◆  Implementation of weak entities

© 2014 Zvi M. Kedem 139

Key Ideas

◆  Foreign keys ◆  Primary key / foreign key constraints inducing many-to-

one relationships between tables ◆  Concept of referential integrity ◆  Crow’s feet notation: ends of lines ◆  Crow’s feet notation: pattern of lines

Documents

Unit 3 The Relational Model And From ER Diagram To Relational … · 2019-05-07 · The given ER diagram is clear, other than • Discovered, which is the continent in which a particular